If $a^2 +3b^2$ is a prime integer, we have to show that $a +bi\sqrt{3}$ is an irreducible element in $\mathbb{Z}[i\sqrt{3}]$.
Now $a^2+3b^2=(a+ib\sqrt{3})(a-ib\sqrt{3})$ as $a^2+3b^2$ is prime, so $a^2+3b^2 | (a+ib\sqrt{3})$ or $a^2+3b^2 | (a-ib\sqrt{3})$. Suppose $a^2+3b^2 | (a+ib\sqrt{3})$ so $(a+ib\sqrt{3})=q.(a^2+3b^2)$. Then $a^2+3b^2=q.(a^2+3b^2)(a-ib\sqrt{3}) \implies 1=q.(a-ib\sqrt{3}) \implies q \ \text{is a unit}$. So $a+ib\sqrt{3}$ is irreducible.
But if $a^2+3b^2 | (a-ib\sqrt{3})$ so $(a-ib\sqrt{3})=q_1.(a^2+3b^2)$. Then $a^2+3b^2=q_1.(a^2+3b^2)(a+ib\sqrt{3}) \implies 1=q_1.(a+ib\sqrt{3}) \implies q_1 \ \text{is a unit}$. From this how to show $(a+bi\sqrt{3})$ is irreducible.
And is there exists any prime integer in $\mathbb{Z}[i\sqrt{3}]$ of the form $a^2+3b^2$?